Question

If $f\left(x\right)$ is continuous at $x=0$, where $f\left(x\right)=\frac{8^x-2^x}{k^k-1}$, for $x\ne 0$
$=12$, for $x=0$
Find $k$.

Answer 1

Given
$\frac{8^x-2^x}{k^k-1}$ for $x$ $=0$
$f\left(x\right)=12$ for x $\ne 0$
Since $f\left(x\right)$ is continuous at $x=0$
$\therefore \underset{x\to 0}{lim}$ $f\left(x\right)=12$
$\underset{x\to 0}{lim}\frac{8^x-2^x}{k^k-1}=12$
Or $\underset{x\to 0}{lim}{8}^x-2^x=12\left(k^k-1\right)$
Or $1-1=$$12\left(k^k-1\right)$
Or $0=$$12\left(k^k-1\right)$
Or $0=$ $k^k-1$
$\therefore k^k=1$
Now taking log both sides
$log{k}^k=\log 1$
$k\log k=0$
Therefore either $k=0$ or $\log k=0$
But if $k=0$ , then logk can not be defined.
So $\log k$ must be $0$ which gives $k=1$.